Find a point on the curve $y=(x-2)^{2}$ at which the tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$.

The equation of the tangent to the curve $y=4xe^{x}$ at the point $\left(-1, -\frac{4}{e}\right)$ is:

Let $f: R \rightarrow R$ be a bijection. $A$ curve represented by $y=f(x)$ is such that $f^{\prime}(x)>0$ for all $x \in R$. The tangent and normal drawn at $P(\alpha, 1)$ on the curve cut the $X$-axis at $A$ and $B$ respectively, and $C$ is the foot of the perpendicular from $P$ onto the $X$-axis. If $P(\alpha, 1)$ is such a point that $AC+CB$ is minimum, then the tangent at $P$ is parallel to the line

The equation of the tangent to the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$ $(n \in N)$ at the point with abscissa equal to '$a$' is:

Find the equations of the tangent and normal to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$ at $(1,1).$

The number of points on the curve $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$ is: